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 n, m, x[100005];
set<pair<int, int> > w[32];
long long r;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) scanf("%d", x + i);
  for (int k = 0, ThxDem = 17; k < ThxDem; ++k) {
    int i = 0;
    while (i < n) {
      if (!((x[i] >> k) & 1)) {
        i++;
        continue;
      }
      int j = i;
      while (j < n && ((x[j] >> k) & 1)) j++;
      w[k].insert(make_pair(i, j));
      r += 1LL * (j - i) * (j - i + 1) / 2 * (1 << k);
      i = j;
    }
  }
  while (m--) {
    int i, v;
    scanf("%d%d", &i, &v);
    i--;
    for (int k = 0, ThxDem = 17; k < ThxDem; ++k)
      if (((x[i] >> k) & 1) != ((v >> k) & 1)) {
        if (((v >> k) & 1)) {
          int a = i, b = i + 1;
          vector<pair<int, int> > d;
          auto it = w[k].lower_bound(make_pair(i, -1));
          if (it != w[k].end() && it->first == i + 1) {
            r -= 1LL * (it->second - it->first) * (it->second - it->first + 1) /
                 2 * (1 << k);
            b = it->second;
            d.push_back(*it);
          }
          if (it != w[k].begin() && (--it)->second == i) {
            r -= 1LL * (it->second - it->first) * (it->second - it->first + 1) /
                 2 * (1 << k);
            a = it->first;
            d.push_back(*it);
          }
          for (auto p : d) w[k].erase(p);
          r += 1LL * (b - a) * (b - a + 1) / 2 * (1 << k);
          w[k].insert(make_pair(a, b));
        } else {
          auto it = w[k].lower_bound(make_pair(i + 1, -1));
          --it;
          r -= 1LL * (it->second - it->first) * (it->second - it->first + 1) /
               2 * (1 << k);
          auto p = *it;
          w[k].erase(it);
          if (p.first != i) {
            r += 1LL * (i - p.first) * (i - p.first + 1) / 2 * (1 << k);
            w[k].insert(make_pair(p.first, i));
          }
          if (p.second != i + 1) {
            r += 1LL * (p.second - (i + 1)) * (p.second - (i + 1) + 1) / 2 *
                 (1 << k);
            w[k].insert(make_pair(i + 1, p.second));
          }
        }
      }
    x[i] = v;
    printf("%lld\n", r);
  }
  return 0;
}