You are given circular array a0, a1, ..., an - 1. 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.

Input

The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a0, a1, ..., an - 1 ( - 106 ≤ ai ≤ 106), ai 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; - 106 ≤ v ≤ 106) — inc operation.

Output

For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1


Output

1
0
0
import java.util.*;
import java.io.*;
import java.util.Map.Entry;

public class Codeforces {

    public static class SegmentTree {        // 1-based DS, OOP

        int N;            //the number of elements in the array as a power of 2 (i.e. after padding)
        long[] array, sTree, lazy;

        SegmentTree(long[] in) {
            array = in;
            N = in.length - 1;
            sTree = new long[N << 1];        //no. of nodes = 2*N - 1, we add one to cross out index zero
            lazy = new long[N << 1];
            Arrays.fill(sTree, Long.MAX_VALUE);
            build(1, 1, N);
        }

        void build(int node, int b, int e)    // O(n)
        {
            if (b == e)
                sTree[node] = array[b];
            else {
                int mid = b + e >> 1;
                build(node << 1, b, mid);
                build(node << 1 | 1, mid + 1, e);
                sTree[node] = Math.min(sTree[node << 1], sTree[node << 1 | 1]);
            }
        }


        void update_point(int index, int val)            // O(log n)
        {
            index += N - 1;
            sTree[index] += val;
            while (index > 1) {
                index >>= 1;
                sTree[index] = Math.min(sTree[index << 1], sTree[index << 1 | 1]);
            }
        }


        void update_range(int i, int j, int val)        // O(log n)
        {
            update_range(1, 1, N, i, j, val);
        }

        void update_range(int node, int b, int e, int i, int j, int val) {
            if (i > e || j < b)
                return;
            if (b >= i && e <= j) {
                sTree[node] += val;
                lazy[node] += val;
            } else {
                int mid = b + e >> 1;
                propagate(node, b, mid, e);
                update_range(node << 1, b, mid, i, j, val);
                update_range(node << 1 | 1, mid + 1, e, i, j, val);
                sTree[node] = Math.min(sTree[node << 1], sTree[node << 1 | 1]);
            }

        }

        void propagate(int node, int b, int mid, int e) {
            lazy[node << 1] += lazy[node];
            lazy[node << 1 | 1] += lazy[node];
            sTree[node << 1] += lazy[node];
            sTree[node << 1 | 1] += lazy[node];
            lazy[node] = 0;
        }

        long query(int i, int j) {
            return query(1, 1, N, i, j);
        }

        long query(int node, int b, int e, int i, int j)    // O(log n)
        {
            if (i > e || j < b)
                return Long.MAX_VALUE;
            if (b >= i && e <= j)
                return sTree[node];
            int mid = b + e >> 1;
            propagate(node, b, mid, e);
            long q1 = query(node << 1, b, mid, i, j);
            long q2 = query(node << 1 | 1, mid + 1, e, i, j);
            return Math.min(q1, q2);

        }
    }

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int n = Integer.parseInt(br.readLine());
        int N = 1;
        while(N < n){
            N <<= 1;
        }
        long[] arr = new long[N + 1];
        StringTokenizer st = new StringTokenizer(br.readLine());
        for(int i = 1; i <= n; i++){
            arr[i] = Integer.parseInt(st.nextToken());
        }
        SegmentTree segmentTree = new SegmentTree(arr);
        int m = Integer.parseInt(br.readLine());
        while(m-- > 0){
            st = new StringTokenizer(br.readLine());
            if(st.countTokens() == 3){
                int lg = Integer.parseInt(st.nextToken());
                int rg = Integer.parseInt(st.nextToken());
                int v = Integer.parseInt(st.nextToken());
                if(lg <= rg){
                    segmentTree.update_range(lg + 1, rg + 1, v);
                }else{
                    segmentTree.update_range(lg + 1, n, v);
                    segmentTree.update_range(1, rg + 1, v);
                }
            }else{
                int lg = Integer.parseInt(st.nextToken());
                int rg = Integer.parseInt(st.nextToken());
                if(lg <= rg){
                    pw.println(segmentTree.query(lg + 1, rg + 1));
                }else{
                    pw.println(Math.min(segmentTree.query(lg + 1, n), segmentTree.query(1, rg + 1)));
                }
            }
        }
        pw.flush();
        pw.close();
    }
}