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;

const int N = 3005;
const int mod = 1e9 + 7;
const int two = 5e8 + 4;

void add(int& x,int y) {
	x += y; if (x >= mod) x -= mod;
}

int mul(int x,int y) {
	return (long long) x * y % mod;
}

int n, q;
int a[N];
int f[N][N];
int nf[N][N];

int main() {
	scanf("%d %d", &n, &q);
	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) {
			f[i][j] = (a[i] > a[j]);
		}
	}
	for (int it = 1; it <= q; ++it) {
		int x, y; 
		scanf("%d %d", &x, &y);
		if (x > y) swap(x, y);
		nf[x][y] = nf[y][x] = mul(two, f[x][y] + f[y][x]);
		for (int i = 1; i <= n; ++i) if (i != x && i != y) {
			nf[i][x] = nf[i][y] = mul(two, f[i][x] + f[i][y]);
			nf[x][i] = nf[y][i] = mul(two, f[x][i] + f[y][i]);
		}
		for (int i = 1; i <= n; ++i) {
			f[i][x] = nf[i][x];
			f[x][i] = nf[x][i];
			f[i][y] = nf[i][y];
			f[y][i] = nf[y][i];
		}
	}
	int ans = 0;
	for (int i = 1; i <= n; ++i) {
		for (int j = i + 1; j <= n; ++j) {
			add(ans, f[i][j]);
		}
	}
	for (int i = 1; i <= q; ++i) {
		ans = mul(ans, 2);
	}
	printf("%d\n", ans);
}