#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
#define int long long

const int N = 2e5 + 5, p = 1e9 + 7;
pair<int, int> nums[N], heap[N];
int ans, arr[N], Logarr[N], dLog[N];

bool smaller(pair<int, int> a, pair<int, int> b)
{
	double x = log2(nums[a.first].first) + nums[a.first].second;
	double y = log2(nums[b.first].first) + nums[b.first].second;
	if (x < y) return true; else return false;
}

void heapInsert(int idx)
{
	while (idx != 0 && smaller(heap[idx], heap[idx / 2]))
	{
		pair<int, int> tmp = heap[idx]; heap[idx] = heap[idx / 2];
		heap[idx / 2] = tmp; idx = idx / 2;
	}
}

void heapify(int idx, int size)
{
	int l = 2 * idx;
	while (l <= size)
	{
		int mini = l + 1 <= size && smaller
			(heap[l + 1], heap[l]) ? l + 1 : l;
		if (smaller(heap[idx], heap[mini])) break;
		pair<int, int> tmp = heap[mini]; heap[mini] = heap[idx];
		heap[idx] = tmp; idx = mini; l = 2 * idx;
	}
}

int power(int a, int b)
{
	int tmp = 2, ans = 1;
	while (b > 0)
	{
		if (b % 2) ans = (ans * tmp) % p;
		tmp = (tmp * tmp) % p; b /= 2;
	}
	return (a * ans) % p;
}

signed main()
{
	int n, k; cin >> n >> k;
	for (int i = 1; i <= n; i++)
	{
		cin >> nums[i].first;
		nums[i].second = 0;
		heap[i] = { i, nums[i].first };
		heapInsert(i);
	}

	for (int i = 0; i < k; i++)
	{
		//时间复杂度O(k*logn)，k比较大时无法在时限内得到结果
		heap[1].second = (heap[1].second * 2) % p;
		nums[heap[1].first].second++; heapify(1, n);
	}

	for (int i = 1; i <= n; i++)
	{
		ans += heap[i].second; ans %= p;
	}
	cout << ans;

	return 0;
}