Problem: Bankopolis, the city you already know, finally got a new bank opened! Unfortunately, its security system is not yet working fine... Meanwhile hacker Leha arrived in Bankopolis and decided to test the system!

Bank has n cells for clients' money. A sequence from n numbers a1, a2, ..., an describes the amount of money each client has. Leha wants to make requests to the database of the bank, finding out the total amount of money on some subsegments of the sequence and changing values of the sequence on some subsegments. Using a bug in the system, Leha can requests two types of queries to the database:

  * 1 l r x y denoting that Leha changes each digit x to digit y in each element of sequence ai, for which l ≤ i ≤ r is holds. For example, if we change in number 11984381 digit 8 to 4, we get 11944341. It's worth noting that Leha, in order to stay in the shadow, never changes digits in the database to 0, i.e. y ≠ 0. 
  * 2 l r denoting that Leha asks to calculate and print the sum of such elements of sequence ai, for which l ≤ i ≤ r holds. 



As Leha is a white-hat hacker, he don't want to test this vulnerability on a real database. You are to write a similar database for Leha to test.

Input

The first line of input contains two integers n and q (1 ≤ n ≤ 105, 1 ≤ q ≤ 105) denoting amount of cells in the bank and total amount of queries respectively. 

The following line contains n integers a1, a2, ..., an (1 ≤ ai < 109) denoting the amount of money in each cell initially. These integers do not contain leading zeros.

Each of the following q lines has one of the formats:

  * 1 l r x y (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 9, 1 ≤ y ≤ 9), denoting Leha asks to change each digit x on digit y for each element ai of the sequence for which l ≤ i ≤ r holds; 
  * 2 l r (1 ≤ l ≤ r ≤ n), denoting you have to calculate and print the sum of elements ai for which l ≤ i ≤ r holds. 

Output

For each second type query print a single number denoting the required sum.

Examples

Input

5 5
38 43 4 12 70
1 1 3 4 8
2 2 4
1 4 5 0 8
1 2 5 8 7
2 1 5


Output

103
207


Input

5 5
25 36 39 40 899
1 1 3 2 7
2 1 2
1 3 5 9 1
1 4 4 0 9
2 1 5


Output

111
1002

Note

Let's look at the example testcase.

Initially the sequence is [38, 43, 4, 12, 70]. 

After the first change each digit equal to 4 becomes 8 for each element with index in interval [1; 3]. Thus, the new sequence is [38, 83, 8, 12, 70].

The answer for the first sum's query is the sum in the interval [2; 4], which equal 83 + 8 + 12 = 103, so the answer to this query is 103.

The sequence becomes [38, 83, 8, 12, 78] after the second change and [38, 73, 7, 12, 77] after the third.

The answer for the second sum's query is 38 + 73 + 7 + 12 + 77 = 207.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int INF = (int)1e9;
const double EPS = 0.000000000000001;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
using namespace std;
void fast_io() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void file_in(const string& filename) { freopen(filename.c_str(), "r", stdin); }
void file_out(const string& filename) {
  freopen(filename.c_str(), "w", stdout);
}
double diffclock(clock_t clock1, clock_t clock2) {
  double diffticks = clock2 - clock1;
  return diffticks / ((double)CLOCKS_PER_SEC / 1000);
}
void sleep(int milliseconds) {
  clock_t start = clock();
  for (clock_t end = clock(); diffclock(start, end) < milliseconds;
       end = clock()) {
  }
}
struct Node {
  long long* digits = new long long[10];
  int* changes = new int[10];
  ~Node() {
    delete[] digits;
    delete[] changes;
  }
  void merge(Node& a, Node& b) {
    for (int i = 0; i < 10; ++i) {
      digits[i] = a.digits[i] + b.digits[i];
    }
  }
  long long count() {
    long long sum = 0;
    for (int i = 0; i < 10; ++i) {
      sum += (long long)digits[i] * i;
    }
    return sum;
  }
};
int* data_arr = new int[100000];
Node* tree = new Node[4 * 100000];
long long* changes_left = new long long[10];
long long* changes_right = new long long[10];
long long* new_digits = new long long[10];
void push(int p, int left, int right) {
  for (int i = 0; i < 10; ++i) {
    new_digits[i] = 0;
  }
  if (right - left > 1) {
    for (int i = 0; i < 10; ++i) {
      changes_left[i] = tree[p].changes[tree[2 * p + 1].changes[i]];
      changes_right[i] = tree[p].changes[tree[2 * p + 2].changes[i]];
    }
    for (int i = 0; i < 10; ++i) {
      tree[2 * p + 1].changes[i] = changes_left[i];
      tree[2 * p + 2].changes[i] = changes_right[i];
    }
  }
  for (int i = 0; i < 10; ++i) {
    new_digits[tree[p].changes[i]] += tree[p].digits[i];
  }
  for (int i = 0; i < 10; ++i) {
    tree[p].digits[i] = new_digits[i];
    tree[p].changes[i] = i;
  }
}
void build(int p, int left, int right) {
  if (right - left == 1) {
    int cur_pow = 1;
    for (int i = 0; i < 10; ++i) {
      tree[p].digits[i] = 0;
      tree[p].changes[i] = i;
    }
    for (int i = 0; (i < 10) && data_arr[left]; ++i) {
      tree[p].digits[data_arr[left] % 10] += cur_pow;
      data_arr[left] /= 10;
      cur_pow *= 10;
    }
    return;
  }
  for (int i = 0; i < 10; ++i) {
    tree[p].changes[i] = i;
  }
  int mid = (left + right) / 2;
  build(2 * p + 1, left, mid);
  build(2 * p + 2, mid, right);
  tree[p].merge(tree[2 * p + 1], tree[2 * p + 2]);
}
void update(int p, int left, int right, int t_left, int t_right, int old_value,
            int new_value) {
  push(p, left, right);
  if ((right <= t_left) || (t_right <= left)) {
    return;
  }
  if ((t_left <= left) && (right <= t_right)) {
    tree[p].changes[old_value] = new_value;
    push(p, left, right);
    return;
  }
  int mid = (left + right) / 2;
  update(2 * p + 1, left, mid, t_left, t_right, old_value, new_value);
  update(2 * p + 2, mid, right, t_left, t_right, old_value, new_value);
  tree[p].merge(tree[2 * p + 1], tree[2 * p + 2]);
}
long long get_sum(int p, int left, int right, int t_left, int t_right) {
  if ((right <= t_left) || (t_right <= left)) {
    return 0;
  }
  push(p, left, right);
  if ((t_left <= left) && (right <= t_right)) {
    return tree[p].count();
  }
  int mid = (left + right) / 2;
  return get_sum(2 * p + 1, left, mid, t_left, t_right) +
         get_sum(2 * p + 2, mid, right, t_left, t_right);
}
signed main() {
  fast_io();
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> data_arr[i];
  }
  build(0, 0, n);
  for (; q > 0; --q) {
    int operation;
    cin >> operation;
    if (operation == 1) {
      int left, right, x, y;
      cin >> left >> right >> x >> y;
      update(0, 0, n, left - 1, right, x, y);
    } else {
      int left, right;
      cin >> left >> right;
      cout << get_sum(0, 0, n, left - 1, right) << '\n';
    }
  }
  return 0;
}