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 <iostream>
#include <cstdio>
#define int long long
using namespace std;
const int N = 3e3 + 5;
const int mod = 1e9 + 7;
int n, Q, f[N][N], a[N], inv2, pow = 1, ans = 0;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
	return x * f;
}
signed main()
{
	n = read(); Q = read();
	for(int i = 1; i <= n; i ++) a[i] = read();
	for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) f[i][j] = (a[i] < a[j]);
	inv2 = (mod + 1) / 2;
	for(int i = 1, x, y; i <= Q; i ++)
	{
		x = read(); y = read(); pow = pow * 2 % mod;
		f[x][y] = f[y][x] = (f[x][y] + f[y][x]) % mod * inv2 % mod;
		for(int j = 1; j <= n; j ++)
		{
			if(j == x || j == y) continue;
			f[x][j] = f[y][j] = (f[x][j] + f[y][j]) % mod * inv2 % mod;
			f[j][x] = f[j][y] = (f[j][x] + f[j][y]) % mod * inv2 % mod;
		}
	}
	for(int i = 1; i <= n; i ++) for(int j = i - 1; j >= 1; j --) ans = (ans + f[i][j]) % mod;
	printf("%lld\n", ans * pow % mod);
	return 0;
}