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>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, Q, tree[1 << 19], s[1 << 18], DP[1 << 18];
pair<int, int>lr[200000];
bool b[1 << 18];

void update(int x, int t) {
	int l, r;
	l = r = x;
	x += 1 << 18;
	tree[x] = min(tree[x], t);
	while (x != 1) {
		if (x & 1) {
			l -= r - l + 1;
		}
		else {
			r += r - l + 1;
		}
		x /= 2;
		tree[x] = min(tree[x * 2] + (r - l + 1) / 2 - (s[r] - s[(l + r) / 2]), tree[x * 2 + 1]);
	}
}

int calc(int l, int r, int p, int q, int k) {
	if (r < p || q < l)return 0xE869120;
	if (p <= l&&r <= q)return tree[k] + q - r - (s[q] - s[r]);
	return min(calc(l, (l + r) / 2, p, q, k * 2), calc((l + r) / 2 + 1, r, p, q, k * 2 + 1));
}

int main() {
	cin >> N;
	for (int i = 1; i <= N; ++i) {
		cin >> b[i];
		s[i] = b[i] + s[i - 1];
	}
	cin >> Q;
	for (int i = 0; i < Q; ++i) {
		cin >> lr[i].first >> lr[i].second;
	}
	sort(lr, lr + Q);
	for (int i = 0; i < 1 << 18; ++i) {
		DP[i] = 0xE869120;
	}
	for (int i = 0; i < 1 << 19; ++i) {
		tree[i] = 0xE869120;
	}
	update(0, 0);
	DP[0] = 0;
	int t = 0;
	for (int i = 1; i <= N; ++i) {
		DP[i] = min(DP[i], DP[i - 1] + b[i]);
		while (t != Q&&lr[t].first == i) {
			DP[lr[t].second] = min({ DP[lr[t].second], DP[lr[t].first - 1] + lr[t].second - lr[t].first + 1 - (s[lr[t].second] - s[lr[t].first - 1]), calc(0, (1 << 18) - 1, lr[t].first, lr[t].second, 1) });
			update(lr[t].second,DP[lr[t].second]);
			++t;
		}
	}
	cout << DP[N] << endl;
}