#include <iostream>
#include <vector>
#include <queue>
#include <cmath>

using namespace std;

const long long MOD = 1e9 + 7;

int main() {
    int n;
    long long k;
    cin >> n >> k;
    vector<long long> a(n);
    
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    // 优先队列，最小堆
    priority_queue<long long, vector<long long>, greater<long long>> minHeap(a.begin(), a.end());
    
    while (k > 0) {
        long long minVal = minHeap.top();
        minHeap.pop();
        minVal *= 2;
        minHeap.push(minVal);
        --k;
        
        // 检查是否所有的数都在 [2^29, 2^30)
        if (minVal >= (1LL << 29)) {
            break;
        }
    }
    
    // 计算所有数的和，并对 MOD 取模
    long long totalSum = 0;
    while (!minHeap.empty()) {
        totalSum = (totalSum + minHeap.top()) % MOD;
        minHeap.pop();
    }
    
    // 剩余的 k 次操作
    if (k > 0) {
        int cycle = n; // 每次循环的长度
        long long fullCycles = k / cycle; // 完整循环的次数
        long long remainder = k % cycle; // 剩余未完成的操作
        
        // 每个数都被乘 2 fullCycles 次
        long long multiplier = 1LL << fullCycles;
        
        // 加上剩余的操作
        for (int i = 0; i < remainder; ++i) {
            a[i] *= 2;
        }
        
        // 计算最终的和，并对 MOD 取模
        totalSum = 0;
        for (int i = 0; i < n; ++i) {
            totalSum = (totalSum + a[i] * multiplier) % MOD;
        }
    }
    
    cout << totalSum << endl;

    return 0;
}