Problem: Range Count Query

Given the sequence a_1, a_2, .., a_N.

In the query, answer the number of terms whose value is l or more and r or less.

input


N Q
a_1 a_2 ... a_N
l_1 r_1
l_2 r_2
::
l_q r_q


output


ans_1
ans_2
::
ans_q


On line i, output the answer to the i-th query, that is, the number of j such as l_i \ leq a_j \ leq r_i.

Constraint

* 1 \ leq N, Q \ leq 10 ^ 5
* 1 \ leq a_i \ leq 10 ^ 9
* 1 \ leq l_i \ leq r_i \ leq 10 ^ 9



Input example


6 3
8 6 9 1 2 1
2 8
1 7
3 5


Output example


3
Four
0






Example

Input

6 3
8 6 9 1 2 1
2 8
1 7
3 5


Output

3
4
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000000000000
#define MOD 1000000007
#define all(v) v.begin(),v.end()
#define pb push_back
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int kaijo[2000010];
struct edge{int to,cost;};
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
void init_fact(){
    kaijo[0]=1;
    for(int i=1;i<=2000000;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=MOD;
    }
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%MOD;
    int memo=modpow(a,b/2);
    return memo*memo%MOD;
}
int comb(int a,int b){
    if(!kaijo[0])init_fact();
    return kaijo[a]*modpow(kaijo[a-b],MOD-2)%MOD*modpow(kaijo[b],MOD-2)%MOD;
}
int inv(int x){
    x=modpow(x,MOD-2);
    return x;
}
bool kosa(double ax,double ay,double bx,double by,double cx,double cy,double dx,double dy){
    double ta=(cx-dx)*(ay-cy)+(cy-dy)*(cx-ax);
    double tb=(cx-dx)*(by-cy)+(cy-dy)*(cx-bx);
    double tc=(ax-bx)*(cy-ay)+(ay-by)*(ax-cx);
    double td=(ax-bx)*(dy-ay)+(ay-by)*(ax-dx);
    return tc*td<0&&ta*tb<0;
}
double dist(double ax,double ay,double bx,double by){
    return sqrt((bx-ax)*(bx-ax)+(ay-by)*(ay-by));
}
int n,q,a[100010];
signed main(){
    cin>>n>>q;
    a[0]=0;
    for(int i=1;i<=n;i++)cin>>a[i];
    sort(a,a+n+1);
    rep(_,q){
        int l,r;
        cin>>l>>r;
        cout<<upper_bound(a,a+1+n,r)-lower_bound(a,a+1+n,l)<<endl;
    }
    return 0;
}