
// Problem : F - NRE
// Contest : AtCoder Regular Contest 085
// URL : https://atcoder.jp/contests/arc085/tasks/arc085_d
// Memory Limit : 256 MB
// Time Limit : 3000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define pb push_back
#define mp make_pair
#define y1 y114514
#define y2 y1919810
typedef pair <int, int> PII;
//#define int long long
#define all(x) (x.begin(),  x.end())
#define lsh(i) (1 << (i))
#define lshll(i) (1LL << (i))
#define repn(i, n) for (int i = 1; i <= (int)n; i++)
#define rep(i, n) for (int i = 0; i < (int)n; i++)
#define range(i, s, t)  for (int i = s; i <= (int)t; i++)
const int INF = 0x3f3f3f3f;
const long double EPS = 1e-6;
long long yuri() {
	char c = getchar();
	int neg = 1;
	long long val = 0;
	while (c != '-' && !isdigit(c)) c = getchar();
	if (c == '-') c = getchar(), neg = -1;
	while (isdigit(c)) val = (val << 3) + (val << 1) + c - '0', c = getchar();
	return neg * val;
}
const int N = 262144;
int natsuki[N * 2], n, inv[N], b[N], dp[N];
vector <int> qry[N];
void monika(int l, int r, int x) {
	l += N;
	r += N;
	while (l <= r) {
		natsuki[l] = min(natsuki[l], x); natsuki[r] = min(natsuki[r], x);
		l = (l + 1) >> 1; r = (r - 1) >> 1;
	}
}
int sayori(int l) {
	int r = 1e9;
	l += N;
	while (l) {
		r = min(r, natsuki[l]);
		l >>= 1;
	}
	return r;
}
int main() {
	n = yuri(); repn(i, N * 2 - 1) natsuki[i] = 1e9;
	repn(i, n) {b[i] = yuri();  inv[i] = inv[i - 1] + (1 - b[i]);}
	int q = yuri(); 
	while (q--) {int l = yuri();int r = yuri(); qry[r].pb(l);}
	repn(i, n) {
		monika(i - 1, i - 1, dp[i - 1] - inv[i - 1]);
		dp[i] = dp[i - 1] + (b[i]);
		for (auto v:qry[i]) {
			int l = v, r = i; 
			monika(l, r, sayori(l - 1));
		}
		dp[i] = min(dp[i], sayori(i) + inv[i]);
	}
	cout << dp[n] << endl;
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp

// Problem : F - NRE
// Contest : AtCoder Regular Contest 085
// URL : https://atcoder.jp/contests/arc085/tasks/arc085_d
// Memory Limit : 256 MB
// Time Limit : 3000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define pb push_back
#define mp make_pair
#define y1 y114514
#define y2 y1919810
typedef pair <int, int> PII;
//#define int long long
#define all(x) (x.begin(),  x.end())
#define lsh(i) (1 << (i))
#define lshll(i) (1LL << (i))
#define repn(i, n) for (int i = 1; i <= (int)n; i++)
#define rep(i, n) for (int i = 0; i < (int)n; i++)
#define range(i, s, t)  for (int i = s; i <= (int)t; i++)
const int INF = 0x3f3f3f3f;
const long double EPS = 1e-6;
long long yuri() {
	char c = getchar();
	int neg = 1;
	long long val = 0;
	while (c != '-' && !isdigit(c)) c = getchar();
	if (c == '-') c = getchar(), neg = -1;
	while (isdigit(c)) val = (val << 3) + (val << 1) + c - '0', c = getchar();
	return neg * val;
}
const int N = 262144;
int natsuki[N * 2], n, inv[N], b[N], dp[N];
vector <int> qry[N];
void monika(int l, int r, int x) {
	l += N;
	r += N;
	while (l <= r) {
		natsuki[l] = min(natsuki[l], x); natsuki[r] = min(natsuki[r], x);
		l = (l + 1) >> 1; r = (r - 1) >> 1;
	}
}
int sayori(int l) {
	int r = 1e9;
	l += N;
	while (l) {
		r = min(r, natsuki[l]);
		l >>= 1;
	}
	return r;
}
int main() {
	n = yuri(); repn(i, N * 2 - 1) natsuki[i] = 1e9;
	repn(i, n) {b[i] = yuri();  inv[i] = inv[i - 1] + (1 - b[i]);}
	int q = yuri(); 
	while (q--) {int l = yuri();int r = yuri(); qry[r].pb(l);}
	repn(i, n) {
		monika(i - 1, i - 1, dp[i - 1] - inv[i - 1]);
		dp[i] = dp[i - 1] + (b[i]);
		for (auto v:qry[i]) {
			int l = v, r = i; 
			monika(l, r, sayori(l - 1));
		}
		dp[i] = min(dp[i], sayori(i) + inv[i]);
	}
	cout << dp[n] << endl;
	return 0;
}

```