You are given circular array *a*<sub class="lower-index">0</sub>, *a*<sub class="lower-index">1</sub>, ..., *a*<sub class="lower-index">*n* - 1</sub>. There are two types of operations with it: 
 -  *inc*(*lf*, *rg*, *v*) — this operation increases each element on the segment [*lf*, *rg*] (inclusively) by *v*; -  *rmq*(*lf*, *rg*) — this operation returns minimal value on the segment [*lf*, *rg*] (inclusively). 
Assume segments to be circular, so if *n* = 5 and *lf* = 3, *rg* = 1, it means the index sequence: 3, 4, 0, 1.

Write program to process given sequence of operations.
The first line contains integer *n* (1 ≤ *n* ≤ 200000). The next line contains initial state of the array: *a*<sub class="lower-index">0</sub>, *a*<sub class="lower-index">1</sub>, ..., *a*<sub class="lower-index">*n* - 1</sub> ( - 10<sup class="upper-index">6</sup> ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">6</sup>), *a*<sub class="lower-index">*i*</sub> are integer. The third line contains integer *m* (0 ≤ *m* ≤ 200000), *m* — the number of operartons. Next *m* lines contain one operation each. If line contains two integer *lf*, *rg* (0 ≤ *lf*, *rg* ≤ *n* - 1) it means *rmq* operation, it contains three integers *lf*, *rg*, *v* (0 ≤ *lf*, *rg* ≤ *n* - 1; - 10<sup class="upper-index">6</sup> ≤ *v* ≤ 10<sup class="upper-index">6</sup>) — *inc* operation.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

typedef long long ll;
const int N=100005;
int f[N<<2],lazy[N<<2];
void build (int root,int left,int right){
	lazy[root]=0;
	if(left==right){
		scanf("%d",&f[root]);
		return ;
	}
	int mid=(left+right)>>1;
	int rt=root<<1;
	build(rt,left,mid);
	build(rt+1,mid+1,right);
	f[root]=min(f[rt],f[rt+1]);
}
void pushdown(int root,int left,int right){
	if(lazy[root]){
		int rt=root<<1;
		lazy[rt]+=lazy[root];
		lazy[rt+1]+=lazy[root];
		f[rt]+=lazy[root];
		f[rt+1]+=lazy[root];
		lazy[root]=0; 
	}
}
void update(int root,int left,int right,int uleft,int uright,int val){
	if(uleft<=left&&right<=uright){
		lazy[root]+=val;
		f[root]+=val;
		return ;
	}
	pushdown(root,left,right);
	int mid=(left+right)>>1;
	int rt=root<<1;
	if(uleft<=mid)update(rt,left,mid,uleft,uright,val);
	if(mid<uright)update(rt+1,mid+1,right,uleft,uright,val);
    f[root]=min(f[rt],f[rt+1]);
}
ll query(int root,int left,int right,int qleft,int qright){
	ll ans=1000000000;
	if(qleft<=left&&right<=qright){
		return f[root];
	}
	pushdown(root,left,right);
	int rt=root<<1;
	int mid=(left+right)>>1;
	 
	if(qleft<=mid)ans=min(ans,query(rt,left,mid,qleft,qright));
	if(mid<qright)ans=min(ans,query(rt+1,mid+1,right,qleft,qright));
	return ans;
}
int main(){
	int n,m;
	int a,b;
	ll c;
	char ch;
	scanf("%d",&n);
	build(1,1,n);
	scanf("%d",&m);
	while(m>0){
		scanf("%d%d%c",&a,&b,&ch);
		if(ch==' '){
			scanf("%lld",&c);
			if(a>b){
				update(1,1,n,1,b+1,c);
				update(1,1,n,a+1,n,c);
			}else{
				update(1,1,n,a+1,b+1,c);
			}
		}else{
			if(a>b)printf("%d\n",min(query(1,1,n,1,b+1),query(1,1,n,a+1,n)));
			else printf("%d\n",query(1,1,n,a+1,b+1));
		}
		m--;
	}
	return 0;
}
     							 	   	   		 							