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>
using namespace std;

typedef long long LL;

const int N = 3010, mod = 1e9 + 7;

LL power(LL a, LL b)
{
	LL c = 1;
	for (; b; a = a * a % mod, b >>= 1)
		if (b & 1) c = c * a % mod;
	return c;
}

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

LL inv2 = power(2, mod - 2);
LL f[N][N], g[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 i = 1; i <= q; ++ i)
	{
		scanf("%d %d", &x, &y);
		for (int j = 1; j <= n; ++ j)
		{
			g[x][j] = (f[y][j] * inv2 + f[x][j] * inv2) % mod;
			g[j][x] = (f[j][y] * inv2 + f[j][x] * inv2) % mod;
			
			g[y][j] = (f[x][j] * inv2 + f[y][j] * inv2) % mod;
			g[j][y] = (f[j][x] * inv2 + f[j][y] * inv2) % mod;
		}
		g[x][y] = (f[x][y] * inv2 + f[y][x] * inv2) % mod;
		g[y][x] = (f[y][x] * inv2 + f[x][y] * inv2) % mod;
		for (int j = 1; j <= n; ++ j)
		{
			f[x][j] = g[x][j], f[j][x] = g[j][x];
			f[y][j] = g[y][j], f[j][y] = g[j][y];
		}
	}
	LL ans = 0;
	for (int i = 1; i < n; ++ i)
		for (int j = i + 1; j <= n; ++ j)
			(ans += f[i][j]) %= mod;
	printf("%lld\n", ans * power(2, q) % mod);
}