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:
#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
#include<time.h>
#include<cassert>

using namespace std;
#define ll long long
#define ld long double
#define int long long
#define all(a) (a).begin(),(a).end()
#define EPS (1e-13)
#define fs first
#define sc second
#define xx first
#define yy second.first
#define zz second.second
#define H pair<int, int>
#define P pair<int, pair<int, int>>
#define Q(i,j,k) mkp(i,mkp(j,k))
#define rep(i,n) for(int (i) = 0 ; (i) < (n) ; (i)++)
#define mkp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())
#define ssp(i,n) (i==n-1?"\n":" ")
const int mod = 1e9 + 7;
const int Mod = 998244353;
const ll Inf = 3 * 1e18;
const int inf = 15 * 1e8;
ll read() {
	ll u, k = scanf("%lld", &u);
	return u;
}
ll gcd(ll i, ll j) {
	if (i > j) swap(i, j);
	if (i == 0) return j;
	return gcd(j % i, i);
}
ll mod_pow(ll x, ll n, ll p) {
	ll res = 1;
	while (n > 0) {
		if (n & 1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}
ll bcount(ll x) {
	int sum = 0;
	for (int i = 0; i < 20; i++)
		if ((1ll << i) & x) sum++;
	return sum;
}
/*const int fn_ = 2000000;
ll fact_[fn_], comp_[fn_];
ll comb(ll x, ll y, ll Mod = mod) {
	if (!fact_[0]) {
		fact_[0] = 1; comp_[0] = 1;
		for (int i = 1; i < fn_; i++) {
			fact_[i] = fact_[i - 1] * i % Mod;
			comp_[i] = mod_pow(fact_[i], Mod - 2, Mod);
		}
	}
	if (x < y) return 0;
	return fact_[x] * comp_[x - y] % Mod * comp_[y] % Mod;
}*/
//---------------------------------------------------
signed main() {
	int n, q;
	cin >> n >> q;
	vector<int>a;
	rep(i, n) a.push_back(read());
	sort(all(a));
	rep(i, q) {
		int l, r;
		cin >> l >> r;
		cout << upper_bound(all(a), r) - lower_bound(all(a), l) << endl;
	}
}