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 <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <assert.h>
#include <fstream>

using namespace std;

typedef long long ll;

const int MAXN = 3005;
const ll MOD = 1000000007;

int n,q;
int a[MAXN];

ll ans;
ll f[MAXN][MAXN];

int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> q;
	for (int i = 1;i <= n;i++)
		cin >> a[i];
	for (int i = 1;i <= n;i++)
		for (int j = 1;j <= n;j++)
			if (a[i] > a[j])
				f[i][j] = 1;
	for (int x,y,i = 1;i <= q;i++)
	{
		cin >> x >> y;
		for (int j = 1;j <= n;j++)
		{
			if (j == x || j == y)
				continue;
			f[j][x] = f[j][y] = (f[j][x] + f[j][y]) * ((MOD + 1) / 2) % 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 + 1) / 2) % MOD;
		}
		f[x][y] = f[y][x] = (f[x][y] + f[y][x]) * ((MOD + 1) / 2) % MOD;
	}
	for (int i = 1;i <= n;i++)
		for (int j = i + 1;j <= n;j++)
			(ans += f[i][j]) %= MOD;
	for (int i = 1;i <= q;i++)
		(ans *= 2) %= MOD;
	cout << ans << endl;
	return 0;
}