Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

3
1 0 1
1
1 3

Sample Output 1

1

If you choose to perform the operation, a will become \{1, 1, 1\}, for a hamming distance of 1.

Sample Input 2

3
1 0 1
2
1 1
3 3

Sample Output 2

0

If both operations are performed, a will become \{1, 0, 1\}, for a hamming distance of 0.

Sample Input 3

3
1 0 1
2
1 1
2 3

Sample Output 3

1

Sample Input 4

5
0 1 0 1 0
1
1 5

Sample Output 4

2

It may be optimal to perform no operation.

Sample Input 5

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

Sample Output 5

3

Sample Input 6

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

Sample Output 6

5

Sample Input 7

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

Sample Output 7

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

signed main() {
  int n; cin >> n;
  vector<int> b;
  for (int i=0; i<n; ++i) {
    int x; cin >> x;
    b.emplace_back(x);
  }
  int num_q; cin >> num_q;
  vector<pair<int, int> > q;
  while (num_q--) {
    int a, b; cin >> a >> b;
    q.emplace_back(a-1, b);
  }

  vector<int> b_prefix(n+1);
  partial_sum(b.begin(), b.end(), b_prefix.begin()+1);

  sort(q.begin(), q.end(), [&](pair<int, int> a,
			       pair<int, int> b){
	 return a.second < b.second;
       });

  auto num_zeros = [&b_prefix](int i, int j) {
    return j-i - (b_prefix[j] - b_prefix[i]);
  };
  
  auto it = q.begin(), it_end=q.end();
  vector<int> dp(n+1);
  for (int i=1; i<=n; ++i) {
    dp[i] = dp[i-1] + b[i-1];
    for (; it!=it_end && it->second == i; ++it) {
      // cout << "dp["<<i<<"] for "<<it->first<<"--"<<it->second<<": "
      // 	   << dp[it->first]<<" + "<<num_zeros(it->first+1, i)<<'\n';
      dp[i] = min(dp[i], dp[it->first] + num_zeros(it->first, i));
    }
  }
  cout << dp[n] << '\n';
}