Problem: Inna is fed up with jokes about female logic. So she started using binary logic instead.

Inna has an array of n elements a1[1], a1[2], ..., a1[n]. Girl likes to train in her binary logic, so she does an exercise consisting of n stages: on the first stage Inna writes out all numbers from array a1, on the i-th (i ≥ 2) stage girl writes all elements of array ai, which consists of n - i + 1 integers; the k-th integer of array ai is defined as follows: ai[k] = ai - 1[k] AND ai - 1[k + 1]. Here AND is bit-wise binary logical operation.

Dima decided to check Inna's skill. He asks Inna to change array, perform the exercise and say the sum of all <image> elements she wrote out during the current exercise.

Help Inna to answer the questions!

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — size of array a1 and number of Dima's questions. Next line contains n integers a1[1], a1[2], ..., a1[n] (0 ≤ ai ≤ 105) — initial array elements.

Each of next m lines contains two integers — Dima's question description. Each question consists of two integers pi, vi (1 ≤ pi ≤ n; 0 ≤ vi ≤ 105). For this question Inna should make a1[pi] equals vi, and then perform the exercise. Please, note that changes are saved from question to question.

Output

For each question print Inna's answer on a single line.

Examples

Input

3 4
1 1 1
1 1
2 2
3 2
1 2


Output

6
4
7
12

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  set<int> degree[20];
  cin >> n >> m;
  vector<int> a(n + 1);
  set<int>::iterator it;
  for (int j = 0; j < 20; j++) {
    degree[j].insert(0);
    degree[j].insert(n + 1);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    for (int j = 0; j < 20; j++)
      if ((a[i] & (1 << j)) == 0) degree[j].insert(i);
  }
  long long int sum = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 20; j++) {
      if (a[i] & (1 << j)) {
        it = degree[j].upper_bound(i);
        long long int pos = *it;
        sum += (pos - i) * (1 << j);
      }
    }
  }
  for (int k = 0; k < m; k++) {
    int i, x;
    cin >> i >> x;
    for (int j = 0; j < 20; j++) {
      if ((a[i] & (1 << j)) == 0 && (x & (1 << j))) {
        it = degree[j].find(i);
        it++;
        long long int posfin = *it;
        it--;
        it--;
        long long int posini = *it;
        sum += (posfin - i) * (1 << j) * (i - posini);
        degree[j].erase(i);
      } else if ((a[i] & (1 << j)) && (x & (1 << j)) == 0) {
        degree[j].insert(i);
        it = degree[j].find(i);
        it++;
        long long int posfin = *it;
        it--;
        it--;
        long long int posini = *it;
        sum -= (posfin - i) * (1 << j) * (i - posini);
      }
    }
    a[i] = x;
    cout << sum << endl;
  }
}