Codeforces 746G(构造)

                                                                                                  G. New Roads

                                                                                               time limit per test

                                                                                                   2 seconds

                                                                                           memory limit per test

                                                                                                256 megabytes

                                                                                                          input

                                                                                                  standard input

                                                                                                          output

                                                                                                 standard output

There are n cities in Berland, each of them has a unique id — an integer from1 ton, the capital is the one with id1. Now there is a serious problem in Berland with roads — there are no roads.

That is why there was a decision to build n - 1 roads so that there will be exactly one simple path between each pair of cities.

In the construction plan t integers a₁, a₂, ..., a_t were stated, wheret equals to the distance from the capital to the most distant city, concerning new roads.a_i equals the number of cities which should be at the distancei from the capital. The distance between two cities is the number of roads one has to pass on the way from one city to another.

Also, it was decided that among all the cities except the capital there should be exactlyk cities with exactly one road going from each of them. Such cities are dead-ends and can't be economically attractive. In calculation of these cities the capital is not taken into consideration regardless of the number of roads from it.

Your task is to offer a plan of road's construction which satisfies all the described conditions or to inform that it is impossible.

Input

The first line contains three positive numbers n,t andk (2 ≤ n ≤ 2·10⁵,1 ≤ t, k < n) — the distance to the most distant city from the capital and the number of cities which should be dead-ends (the capital in this number is not taken into consideration).

The second line contains a sequence of t integersa₁, a₂, ..., a_t (1 ≤ a_i < n), thei-th number is the number of cities which should be at the distancei from the capital. It is guaranteed that the sum of all the valuesa_i equalsn - 1.

Output

If it is impossible to built roads which satisfy all conditions, print -1.

Otherwise, in the first line print one integer n — the number of cities in Berland. In the each of the nextn - 1 line print two integers — the ids of cities that are connected by a road. Each road should be printed exactly once. You can print the roads and the cities connected by a road in any order.

If there are multiple answers, print any of them. Remember that the capital has id1.

Examples

Input

7 3 3
2 3 1

Output

7
1 3
2 1
2 6
2 4
7 4
3 5

Input

14 5 6
4 4 2 2 1

Output

14
3 1
1 4
11 6
1 2
10 13
6 10
10 12
14 12
8 4
5 1
3 7
2 6
5 9

Input

3 1 1
2

Output

-1

在构造树的时候，先把树的主链确定，再确定哪些节点为叶子节点(显然深度最大的那些点一定是叶子结点,且根节点一定不是叶子结点因为n≥2)，哪些不是叶子节点。

当叶子节点数目不够时，考虑那些不一定是叶子节点的节点（即深度不是最大值并且不是树的主链的成员的节点),把他作为深度大于他们的结点的父亲即可。这样该结点就变成非叶子结点了。

当非叶子结点个数大于那些可以变成非叶子结点的个数时,无解。

#include <bits/stdc++.h>

using namespace std;

#define REP(i,n)                for(int i(0); i <  (n); ++i)
#define rep(i,a,b)              for(int i(a); i <= (b); ++i)
#define PB      		push_back

const int N     =    200000      +       10;
vector <int> v[N];
int fa[N], a[N], n, la, leaf, cnt, l;

int main(){

	scanf("%d%d%d", &n, &la, &leaf);
	rep(i, 1, la) scanf("%d", a + i);a[0] = 1; 
	if ((a[la] > leaf) || (n - la < leaf) || (n < leaf)){ puts("-1"); return 0;}

	int sum = 1; rep(i, 1, la) sum += a[i];
	if (sum != n){ puts("-1"); return 0;}
	cnt = 0; rep(i, 0, la) rep(j, 1, a[i]) v[i].PB(++cnt);

	REP(i, a[1]) fa[v[1][i]] = 1;
	rep(i, 2, la) fa[v[i][0]] = v[i - 1][0];
	l = n - leaf - la;

	rep(i, 2, la){
		rep(j, 1, a[i] - 1) if (l && j <= a[i - 1] - 1) fa[v[i][j]] = v[i - 1][j], --l;
		else fa[v[i][j]] = v[i - 1][0];
	}

	if (l) {puts("-1"); return 0;}

	printf("%d\n", n);
	rep(i, 2, n) printf("%d %d\n", fa[i], i);
	
	
	return 0;

}