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>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i <  B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
using namespace std;

const int mo = 1e9 + 7;

ll ksm(ll x, ll y) {
	ll s = 1;
	for(; y; y /= 2, x = x * x % mo)
		if(y & 1) s = s * x % mo;
	return s;
}

const ll ni2 = ksm(2, mo - 2);

const int N = 3005;

int n, q, a[N], x[N], y[N];
ll f[N][N];

int main() {
	scanf("%d %d", &n, &q);
	fo(i, 1, n) scanf("%d", &a[i]);
	fo(i, 1, n) fo(j, 1, n) f[i][j] = a[i] > a[j];
	fo(i, 1, q) {
		scanf("%d %d", &x[i], &y[i]);
		fo(j, 1, n) if(j != x[i] && j != y[i]) {
			f[j][y[i]] = f[j][x[i]] = (f[j][y[i]] + f[j][x[i]]) * ni2 % mo;
			f[y[i]][j] = f[x[i]][j] = (f[y[i]][j] + f[x[i]][j]) * ni2 % mo;
		}
		f[x[i]][y[i]] = f[y[i]][x[i]] = (f[x[i]][y[i]] + f[y[i]][x[i]]) * ni2 % mo;
	}
	ll w = ksm(2, q), ans = 0;
	fo(i, 1, n) fo(j, i + 1, n) ans = (ans + f[i][j] * w) % mo;
	pp("%lld\n", ans);
}