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 <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 3005
#define ll long long
#define mod 1000000007
#define inv2 500000004
void Read(int &p)
{
	p = 0;
	char c = getchar();
	for (; c < '0' || c > '9'; c = getchar());
	for (; c >= '0' && c <= '9'; c = getchar())p = p * 10 + c - '0';
}
int n, q;
int f[N][N], A[N];
int main()
{
	Read(n), Read(q);
	for (int i = 1; i <= n; i++)
		Read(A[i]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			f[i][j] = (A[i] > A[j]);
	int ans = 1;
	for (; q--; )
	{
		ans = (ans << 1) % mod;
		int x, y;
		Read(x), Read(y);
		for (int i = 1; i <= n; i++)
		{
			if (i != x && i != y)
			{
				f[i][x] = f[i][y] = (ll)(f[i][x] + f[i][y]) * inv2 % mod;
				f[x][i] = f[y][i] = (ll)(f[x][i] + f[y][i]) * inv2 % mod;
			}
		}
		f[x][y] = f[y][x] = (ll)(f[x][y] + f[y][x]) * inv2 % mod;
	}
	int sum = 0;
	for (int i = 1; i <= n; i++)
		for (int j = i + 1; j <= n; j++)
			sum = (sum + f[i][j]) % mod;
	ans = (ll)sum * ans % mod;
	printf("%d\n", ans);
}
//