Problem: You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

Solution (in CPP):

Solution:
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
//#define INF 1145141919
#define INF 500000
#define MOD 1000000007
#define MAX_N (1<<18)
int seg_m[2*MAX_N-1], seg_a[2*MAX_N-1];

void add(int a, int b, int x, int k=0, int l=0, int r=MAX_N) {
  if (r <= a || b <= l) return;
  if (a <= l && r <= b) {
    seg_a[k] += x;
    return;
  }
  add(a, b, x, k*2+1, l, (l+r)/2);
  add(a, b, x, k*2+2, (l+r)/2, r);
  seg_m[k] = min(
    seg_m[k*2+1] + seg_a[k*2+1],
    seg_m[k*2+2] + seg_a[k*2+2]
  );
}

int query(int a, int b, int k=0, int l=0, int r=MAX_N) {
  if (r <= a || b <= l) return 1145141919;
  if (a <= l && r <= b) return seg_m[k] + seg_a[k];
  return min(query(a, b, k*2+1, l, (l+r)/2),
             query(a, b, k*2+2, (l+r)/2, r)) + seg_a[k];
}
void chmin(int pos, int v) {
  int cur = query(pos, pos+1);
  if (v >= cur) return;
  add(pos, pos+1, v-cur);
}


int N, Q;
int B[200000];
vector<int> G[200000];

signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N;
  rep(i, N) cin >> B[i];
  cin >> Q;
  rep(i, Q) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    G[l].pb(r);
  }
  add(1, MAX_N, INF);
  rep(l, N) {
    for (int r : G[l]) chmin(r+1, query(0, r+2));
    if (B[l]) add(0, l+1, 1);
    else add(l+1, MAX_N, 1);
  }
  int m = query(0, N+1);
  cout << m << "\n";
  return 0;
}