Consider a sequence of digits of length 2^k [a_1, a_2, …, a_{2^k}]. We perform the following operation with it: replace pairs (a_{2i+1}, a_{2i+2}) with (a_{2i+1} + a_{2i+2})mod 10 for 0≤ i<2^{k-1}. For every i where a_{2i+1} + a_{2i+2}≥ 10 we get a candy! As a result, we will get a sequence of length 2^{k-1}.

Less formally, we partition sequence of length 2^k into 2^{k-1} pairs, each consisting of 2 numbers: the first pair consists of the first and second numbers, the second of the third and fourth …, the last pair consists of the (2^k-1)-th and (2^k)-th numbers. For every pair such that sum of numbers in it is at least 10, we get a candy. After that, we replace every pair of numbers with a remainder of the division of their sum by 10 (and don't change the order of the numbers).

Perform this operation with a resulting array until it becomes of length 1. Let f([a_1, a_2, …, a_{2^k}]) denote the number of candies we get in this process. 

For example: if the starting sequence is [8, 7, 3, 1, 7, 0, 9, 4] then:

After the first operation the sequence becomes [(8 + 7)mod 10, (3 + 1)mod 10, (7 + 0)mod 10, (9 + 4)mod 10] = [5, 4, 7, 3], and we get 2 candies as 8 + 7 ≥ 10 and 9 + 4 ≥ 10.

After the second operation the sequence becomes [(5 + 4)mod 10, (7 + 3)mod 10] = [9, 0], and we get one more candy as 7 + 3 ≥ 10. 

After the final operation sequence becomes [(9 + 0) mod 10] = [9]. 

Therefore, f([8, 7, 3, 1, 7, 0, 9, 4]) = 3 as we got 3 candies in total.

You are given a sequence of digits of length n s_1, s_2, … s_n. You have to answer q queries of the form (l_i, r_i), where for i-th query you have to output f([s_{l_i}, s_{l_i+1}, …, s_{r_i}]). It is guaranteed that r_i-l_i+1 is of form 2^k for some nonnegative integer k.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the sequence.

The second line contains n digits s_1, s_2, …, s_n (0 ≤ s_i ≤ 9).

The third line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.

Each of the next q lines contains two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n) — i-th query. It is guaranteed that r_i-l_i+1 is a nonnegative integer power of 2.

Output

Output q lines, in i-th line output single integer — f([s_{l_i}, s_{l_i + 1}, …, s_{r_i}]), answer to the i-th query.

Examples

Input


8
8 7 3 1 7 0 9 4
3
1 8
2 5
7 7


Output


3
1
0


Input


6
0 1 2 3 3 5
3
1 2
1 4
3 6


Output


0
0
1

Note

The first example illustrates an example from the statement.

f([7, 3, 1, 7]) = 1: sequence of operations is [7, 3, 1, 7] → [(7 + 3)mod 10, (1 + 7)mod 10] = [0, 8] and one candy as 7 + 3 ≥ 10 → [(0 + 8) mod 10] = [8], so we get only 1 candy.

f([9]) = 0 as we don't perform operations with it.
import java.io.*;
import java.math.*;
import java.util.*;
import java.util.stream.*;
import java.lang.management.*;
import static java.lang.Math.*;
import static java.lang.Integer.MAX_VALUE;
import static java.lang.Integer.MIN_VALUE;

@SuppressWarnings("unchecked")
public class P1189C {

  final static int SIZE = 1 << 17;

  int [] a;

  class Tree {
    long s = -1;
    int l, r, m;
    Tree lt, rt;

    public Tree(int l, int r) {
      this.l = l;
      this.r = r;
      m = (l + r) >> 1;

      if (l != m) {
        lt = new Tree(l, m);
        rt = new Tree(m, r);
      } else {
        s = (l < a.length) ? a[l] : 0;
      }
    }

    long sum(int l, int r) {
      return (((this.l == l) && (this.r == r)) ? ((s >= 0) ? s : (s = lt.sum(l, m) + rt.sum(m, r)))
                                               : (((l < m) ? lt.sum(l, min(m, r)) : 0) + 
                                                  ((r > m) ? rt.sum(max(l, m), r) : 0)));
    }
  }

  public void run() throws Exception {
    a = readInt(nextInt());
    Tree t = new Tree(0, SIZE);
    for (int q = nextInt(); q > 0; println(t.sum(nextInt() - 1, nextInt()) / 10), q--);
  }

  public static void main(String... args) throws Exception {
    br = new BufferedReader(new InputStreamReader(System.in));
    pw = new PrintWriter(new BufferedOutputStream(System.out));
    new P1189C().run();
    br.close();
    pw.close();
    System.err.println("\n[Time : " + (System.currentTimeMillis() - startTime) + " ms]");
    long gct = 0, gcc = 0;
    for (GarbageCollectorMXBean garbageCollectorMXBean : ManagementFactory.getGarbageCollectorMXBeans()) {
      gct += garbageCollectorMXBean.getCollectionTime();
      gcc += garbageCollectorMXBean.getCollectionCount();
    }
    System.err.println("[GC time : " + gct + " ms, count = " + gcc + "]");
  }

  static long startTime = System.currentTimeMillis();
  static BufferedReader br;
  static PrintWriter pw;
  StringTokenizer stok;

  String nextToken() throws IOException {
    while (stok == null || !stok.hasMoreTokens()) {
      String s = br.readLine();
      if (s == null) { return null; }
      stok = new StringTokenizer(s);
    }
    return stok.nextToken();
  }

  void print(byte b) { print("" + b); }
  void print(int i) { print("" + i); }
  void print(long l) { print("" + l); }
  void print(double d) { print("" + d); }
  void print(char c) { print("" + c); }
  void print(Object o) {
    if (o instanceof int[]) { print(Arrays.toString((int [])o));
    } else if (o instanceof long[]) { print(Arrays.toString((long [])o));
    } else if (o instanceof char[]) { print(Arrays.toString((char [])o));
    } else if (o instanceof byte[]) { print(Arrays.toString((byte [])o));
    } else if (o instanceof short[]) { print(Arrays.toString((short [])o));
    } else if (o instanceof boolean[]) { print(Arrays.toString((boolean [])o));
    } else if (o instanceof float[]) { print(Arrays.toString((float [])o));
    } else if (o instanceof double[]) { print(Arrays.toString((double [])o));
    } else if (o instanceof Object[]) { print(Arrays.toString((Object [])o));
    } else { print("" + o); }
  }
  void printsp(int [] a) { for (int i = 0, n = a.length; i < n; print(a[i] + " "), i++); }
  void print(String s) { pw.print(s); }
  void println() { println(""); }
  void println(byte b) { println("" + b); }
  void println(int i) { println("" + i); }
  void println(long l) { println("" + l); }
  void println(double d) { println("" + d); }
  void println(char c) { println("" + c); }
  void println(Object o) { print(o); println(); }
  void println(String s) { pw.println(s); }
  int nextInt() throws IOException { return Integer.parseInt(nextToken()); }
  long nextLong() throws IOException { return Long.parseLong(nextToken()); }
  double nextDouble() throws IOException { return Double.parseDouble(nextToken()); }
  char nextChar() throws IOException { return (char) (br.read()); }
  String next() throws IOException { return nextToken(); }
  String nextLine() throws IOException { return br.readLine(); }
  int [] readInt(int size) throws IOException {
    int [] array = new int [size];
    for (int i = 0; i < size; i++) { array[i] = nextInt(); }
    return array;
  }
  long [] readLong(int size) throws IOException {
    long [] array = new long [size];
    for (int i = 0; i < size; i++) { array[i] = nextLong(); }
    return array;
  }
  double [] readDouble(int size) throws IOException {
    double [] array = new double [size];
    for (int i = 0; i < size; i++) { array[i] = nextDouble(); }
    return array;
  }
  String [] readLines(int size) throws IOException {
    String [] array = new String [size];
    for (int i = 0; i < size; i++) { array[i] = nextLine(); }
    return array;
  }

  int gcd(int a, int b) {
    if (a == 0) return Math.abs(b); if (b == 0) return Math.abs(a);
    a = Math.abs(a); b = Math.abs(b);
    int az = Integer.numberOfTrailingZeros(a), bz = Integer.numberOfTrailingZeros(b);
    a >>>= az; b >>>= bz;
    while (a != b) {
      if (a > b) { a -= b; a >>>= Integer.numberOfTrailingZeros(a); }
            else { b -= a; b >>>= Integer.numberOfTrailingZeros(b); }
    }
    return (a << Math.min(az, bz));
  }

  long gcd(long a, long b) {
    if (a == 0) return Math.abs(b); if (b == 0) return Math.abs(a);
    a = Math.abs(a); b = Math.abs(b);
    int az = Long.numberOfTrailingZeros(a), bz = Long.numberOfTrailingZeros(b);
    a >>>= az; b >>>= bz;
    while (a != b) {
      if (a > b) { a -= b; a >>>= Long.numberOfTrailingZeros(a); }
            else { b -= a; b >>>= Long.numberOfTrailingZeros(b); }
    }
    return (a << Math.min(az, bz));
  }

  void shuffle(int [] a) { // RANDOM shuffle
    Random r = new Random();
    for (int i = a.length - 1, j, t; i >= 0; j = r.nextInt(a.length), t = a[i], a[i] = a[j], a[j] = t, i--);
  }

  void shuffle(int [] a, int m) { // QUICK shuffle
    for (int i = 0, n = a.length, j = m % n, t; i < n; t = a[i], a[i] = a[j], a[j] = t, i++, j = (i * m) % n);
  }

  void shuffle(long [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      long t = a[i]; a[i] = a[j]; a[j] = t;
    }
  }

  void shuffle(Object [] a) {
    Random r = new Random();
    for (int i = a.length - 1; i >= 0; i--) {
      int j = r.nextInt(a.length);
      Object t = a[i]; a[i] = a[j]; a[j] = t;
    }
  }

  int [] sort(int [] a) {
    final int SHIFT = 16, MASK = (1 << SHIFT) - 1, SIZE = (1 << SHIFT) + 1;
    int n = a.length, ta [] = new int [n], ai [] = new int [SIZE];
    for (int i = 0; i < n; ai[(a[i] & MASK) + 1]++, i++);
    for (int i = 1; i < SIZE; ai[i] += ai[i - 1], i++);
    for (int i = 0; i < n; ta[ai[a[i] & MASK]++] = a[i], i++);
    int [] t = a; a = ta; ta = t;
    ai = new int [SIZE];
    for (int i = 0; i < n; ai[(a[i] >> SHIFT) + 1]++, i++);
    for (int i = 1; i < SIZE; ai[i] += ai[i - 1], i++);
    for (int i = 0; i < n; ta[ai[a[i] >> SHIFT]++] = a[i], i++);
    return ta;
  }

  void flush() {
    pw.flush();
  }

  void pause() {
    flush(); System.console().readLine();
  }
}