#include<bits/stdc++.h>
#define mid ((l+r)>>1)
using namespace std;
const int inf=0x3f3f3f3f;
const int N=200005;
int T[N<<2],tag[N<<2],n,a[N],sum0[N],sum1[N],Q,head;
struct node{int l,r;}q[N];
bool operator < (const node &A,const node &B)
{return A.l<B.l||A.l==B.l&&A.r<B.r;}
void down(int k)
{
  if (tag[k])
  { 
    tag[k<<1]+=tag[k];tag[k<<1|1]+=tag[k];
    T[k<<1]+=tag[k];T[k<<1|1]+=tag[k];
    tag[k]=0;
  }
}
void ins(int k,int l,int r,int x,int y)
{
  if (l==r) {T[k]=min(T[k],y);return;}
  down(k);
  if (x<=mid) ins(k<<1,l,mid,x,y);
  else ins(k<<1|1,mid+1,r,x,y);
  T[k]=min(T[k<<1],T[k<<1|1]);
}
void add(int k,int l,int r,int L,int R,int x)
{
   if (L<=l&&r<=R) {T[k]+=x;tag[k]+=x;return;}
   down(k);
   if (L<=mid) add(k<<1,l,mid,L,R,x);
   if (R>mid) add(k<<1|1,mid+1,r,L,R,x);
   T[k]=min(T[k<<1],T[k<<1|1]);
}
int qry(int k,int l,int r,int L,int R)
{
   if (L<=l&&r<=R) return T[k];
   down(k);int ans=inf;
   if (L<=mid) ans=min(ans,qry(k<<1,l,mid,L,R));
   if (R>mid) ans=min(ans,qry(k<<1|1,mid+1,r,L,R));
   return ans;
}
int main()
{
  scanf("%d",&n);
  for (int i=1;i<=n;i++) 
  {
     scanf("%d",&a[i]);sum0[i]=sum0[i-1];sum1[i]=sum1[i-1];
     if (!a[i]) sum0[i]++;else sum1[i]++;
  }
  scanf("%d",&Q);
  for (int i=1;i<=Q;i++) scanf("%d%d",&q[i].l,&q[i].r);
  sort(q+1,q+Q+1);
  head=1;memset(T,inf,sizeof(T));
  ins(1,0,n,0,0);
  for (int i=1;i<=n;i++)
  {
    while (head<=Q&&q[head].l==i) ins(1,0,n,q[head].r,qry(1,0,n,0,q[head].r)),head++;
    if (a[i]) add(1,0,n,0,i-1,1);else add(1,0,n,0,i-1,-1);
  }
  printf("%d\n",qry(1,0,n,0,n)+sum0[n]);
  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
#include<bits/stdc++.h>
#define mid ((l+r)>>1)
using namespace std;
const int inf=0x3f3f3f3f;
const int N=200005;
int T[N<<2],tag[N<<2],n,a[N],sum0[N],sum1[N],Q,head;
struct node{int l,r;}q[N];
bool operator < (const node &A,const node &B)
{return A.l<B.l||A.l==B.l&&A.r<B.r;}
void down(int k)
{
  if (tag[k])
  { 
    tag[k<<1]+=tag[k];tag[k<<1|1]+=tag[k];
    T[k<<1]+=tag[k];T[k<<1|1]+=tag[k];
    tag[k]=0;
  }
}
void ins(int k,int l,int r,int x,int y)
{
  if (l==r) {T[k]=min(T[k],y);return;}
  down(k);
  if (x<=mid) ins(k<<1,l,mid,x,y);
  else ins(k<<1|1,mid+1,r,x,y);
  T[k]=min(T[k<<1],T[k<<1|1]);
}
void add(int k,int l,int r,int L,int R,int x)
{
   if (L<=l&&r<=R) {T[k]+=x;tag[k]+=x;return;}
   down(k);
   if (L<=mid) add(k<<1,l,mid,L,R,x);
   if (R>mid) add(k<<1|1,mid+1,r,L,R,x);
   T[k]=min(T[k<<1],T[k<<1|1]);
}
int qry(int k,int l,int r,int L,int R)
{
   if (L<=l&&r<=R) return T[k];
   down(k);int ans=inf;
   if (L<=mid) ans=min(ans,qry(k<<1,l,mid,L,R));
   if (R>mid) ans=min(ans,qry(k<<1|1,mid+1,r,L,R));
   return ans;
}
int main()
{
  scanf("%d",&n);
  for (int i=1;i<=n;i++) 
  {
     scanf("%d",&a[i]);sum0[i]=sum0[i-1];sum1[i]=sum1[i-1];
     if (!a[i]) sum0[i]++;else sum1[i]++;
  }
  scanf("%d",&Q);
  for (int i=1;i<=Q;i++) scanf("%d%d",&q[i].l,&q[i].r);
  sort(q+1,q+Q+1);
  head=1;memset(T,inf,sizeof(T));
  ins(1,0,n,0,0);
  for (int i=1;i<=n;i++)
  {
    while (head<=Q&&q[head].l==i) ins(1,0,n,q[head].r,qry(1,0,n,0,q[head].r)),head++;
    if (a[i]) add(1,0,n,0,i-1,1);else add(1,0,n,0,i-1,-1);
  }
  printf("%d\n",qry(1,0,n,0,n)+sum0[n]);
  return 0;
}
```