Codeforces 220B

B. Little Elephant and Array

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai.

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

input

Copy

7 2
3 1 2 2 3 3 7
1 7
3 4

output

Copy

3
1

题目大意：询问区间l,r间有几个数的出现次数等于数值本身。

由于题目满足由区间（l,r）-> (l,r+1)、(l,r-1)、(l-1，r)、(l+1,r)的快速转移，所以可以用莫队算法。刚接触不是太懂，等以后更加了解再更。

#include<bits/stdc++.h>
using namespace std;

const int maxn=100010;
const int maxm=100010;
int n,m,unit;
struct Query{
    int l,r,id;
}node[maxm];

int a[maxn],b[maxn],c[maxn],num[maxn],ans[maxm];

bool cmp(Query a,Query b) {
    if(a.l/unit!=b.l/unit) return a.l/unit<b.l/unit;
    else return a.r<b.r;
}

int main() {
    scanf("%d%d",&n,&m);
    unit=(int)sqrt(n*1.0);
    for(int i=1;i<=n;i++) {
        scanf("%d",&a[i]);
        b[i]=a[i];
    }
    sort(b+1,b+1+n);
    for(int i=1;i<=n;i++) {
        c[i]=lower_bound(b+1,b+1+n,a[i])-b;
    }
    for(int i=1;i<=m;i++) {
        node[i].id=i;
        scanf("%d%d",&node[i].l,&node[i].r);
    }
    sort(node+1,node+m+1,cmp);
    int l=1,r=0,temp=0;
    for(int i=1;i<=m;i++) {
        while(r<node[i].r) { //右扩
            r++;
            if(num[c[r]]+1==a[r]) temp++;
            else if(num[c[r]]==a[r]) temp--;
            num[c[r]]++;
        }
        while(r>node[i].r) { //右缩
            if(num[c[r]]==a[r]) temp--;
            else if(num[c[r]]-1==a[r]) temp++;
            num[c[r]]--;
            r--;
        }
        while(l>node[i].l) { //左扩
            l--;
            if(num[c[l]]+1==a[l]) temp++;
            else if(num[c[l]]==a[l]) temp--;
            num[c[l]]++;
        }
        while(l<node[i].l) {  //左缩
            if(num[c[l]]==a[l]) temp--;
            else if(num[c[l]]-1==a[l]) temp++;
            num[c[l]]--;
            l++;
        }
        ans[node[i].id]=temp;
    }
    for(int i=1;i<=m;i++) printf("%d\n",ans[i]);
}