Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,a,b) for(int i = (b); i --> (a);)
#define all(v) (v).begin(),(v).end()
#define trav(x,v) for(auto &x : v)
#define sz(v) int(v.size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;

const ll mod = 1e9+7;
const ll hlf = 5e8+4;

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);

	int n, q;
	cin >> n >> q;

	vi as(n);
	trav(x, as) cin >> x;

	vector<vector<ll>> mat(n, vector<ll>(n));
	rep(i,0,n) rep(j,0,n) mat[i][j] = (as[i]>as[j]) - (as[j]>as[i]);

	rep(_,0,q){
		int x, y;
		cin >> x >> y;
		--x, --y;
		rep(k,0,n){
			ll res = mat[x][k] + mat[y][k];
			res *= hlf;
			res %= mod;
			mat[x][k] = mat[y][k] = res;
		}
		rep(k,0,n){
			ll res = mat[k][x] + mat[k][y];
			res *= hlf;
			res %= mod;
			mat[k][y] = mat[k][x] = res;
			
		}
	}

	ll ans = 0;

	rep(i,0,n) rep(j,0,i) ans += mat[j][i];
	rep(i,0,n) rep(j,0,i) ans += (as[i]!=as[j]);
	rep(i,0,q) ans = ans * 2 % mod;
	ans = ans * hlf % mod;
	if(ans < 0) ans += mod;

	cout << ans << endl;
}