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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 3003

using namespace std;

const int mod = 1e9 + 7, inv = (mod + 1) / 2;

int n, m;
int a[N], f[N][N], x[N][2], y[N][2];

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i)
		scanf("%d", a + i);
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= n; ++j)
			if(a[i] > a[j]) f[i][j] = 1;
	int sum = 1;
	while(m--){
		int X, Y; (sum *= 2) %= mod;
		scanf("%d%d", &X, &Y);
		for(int i = 1; i <= n; ++i){
			x[i][0] = f[X][i], x[i][1] = f[i][X];
			y[i][0] = f[Y][i], y[i][1] = f[i][Y];
		}
		for(int i = 1; i <= n; ++i){
			f[X][i] = 1ll * inv * (x[i][0] + y[i][0]) % mod;
			f[i][X] = 1ll * inv * (x[i][1] + y[i][1]) % mod;
			f[Y][i] = 1ll * inv * (y[i][0] + x[i][0]) % mod;
			f[i][Y] = 1ll * inv * (y[i][1] + x[i][1]) % mod;
		}
		f[X][Y] = f[Y][X] = 1ll * inv * (x[Y][0] + y[X][0]) % mod;
	}
	int ans = 0;
	for(int i = 1; i <= n; ++i)
		for(int j = i + 1; j <= n; ++j)
			(ans += f[i][j]) %= mod;
	ans = 1ll * ans * sum % mod;
	cout << ans << endl;
	return 0;
}