## Algorithm

D. Changing a String
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

There is a string s, consisting of capital Latin letters. Let's denote its current length as |s|. During one move it is allowed to apply one of the following operations to it:

• INSERT pos ch — insert a letter ch in the string s in the position pos (1 ≤ pos ≤ |s| + 1, A ≤ ch ≤ Z). The letter ch becomes the pos-th symbol of the string s, at that the letters shift aside and the length of the string increases by 1.
• DELETE pos — delete a character number pos (1 ≤ pos ≤ |s|) from the string s. At that the letters shift together and the length of the string decreases by 1.
• REPLACE pos ch — the letter in the position pos of the line s is replaced by ch (1 ≤ pos ≤ |s|, A ≤ ch ≤ Z). At that the length of the string does not change.

Your task is to find in which minimal number of moves one can get a t string from an s string. You should also find the sequence of actions leading to the required results.

Input

The first line contains s, the second line contains t. The lines consist only of capital Latin letters, their lengths are positive numbers from 1 to 1000.

Output

In the first line print the number of moves k in the given sequence of operations. The number should be the minimal possible one. Then print k lines containing one operation each. Print the operations in the format, described above. If there are several solutions, print any of them.

Examples
input
Copy
ABA
ABBBA
output
Copy
2
INSERT 3 B
INSERT 4 B
input
Copy
ACCEPTED
output
Copy
10
REPLACE 1 W
REPLACE 2 R
REPLACE 3 O
REPLACE 4 N
REPLACE 5 G
REPLACE 6 A
INSERT 7 N
INSERT 8 S
INSERT 9 W
REPLACE 11 R

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include <iostream>
#include <string>
#include <algorithm>
#include <memory.h>

using namespace std;

const int N = 1001;
int dp[N][N], shift = 0;
string a, b;

int calc(int i, int j) {
if(i == a.size())
return b.size() - j;

if(j == b.size())
return a.size() - i;

int &res = dp[i][j];
if(res != -1) return res;

res = 0;
if(a[i] == b[j])
return res = calc(i+1, j+1);

return res = min(calc(i+1, j) + 1, min(calc(i, j+1) + 1, calc(i+1, j+1) + 1));
}

void path(int i, int j) {
if(i == a.size()) {
for(int k = j; k < b.size(); k++)
cout << "INSERT " << i+(shift++)+1 << ' ' << b[j+(k-j)] << endl;

return;
};

if(j == b.size()) {
for(int k = i; k < a.size(); k++)
cout << "DELETE " << i+shift+1 << endl;

return;
}

if(a[i] == b[j]) {
path(i+1, j+1);
return;
}

int A = -2, B = -2, C = -2;

A = calc(i+1, j) + 1;
B = calc(i, j+1) + 1;
C = calc(i+1, j+1) + 1;

if(A == calc(i, j)) {
cout << "DELETE " << i+shift+1 << endl;
shift--;
path(i+1, j);
} else if(B == calc(i, j)) {
cout << "INSERT " << i+shift+1 << ' ' << b[j] << endl;
shift++;
path(i, j+1);
} else {
cout << "REPLACE " << i+shift+1 << ' ' << b[j] << endl;
path(i+1, j+1);
}
}

int main() {
cin >> a>> b;

memset(dp, -1, sizeof dp);
cout << calc(0, 0) << endl;
path(0, 0);

return 0;
}
Copy The Code &

Input

cmd
ABA
ABBBA

Output

cmd
2
INSERT 3 B
INSERT 4 B