Problem: A: Find the difference

problem

Given rectangular boards A and B with N squares vertically and M squares horizontally. For each board, each square is painted white or black.

The color of the square in the i-th row and j-th column of the board X is written as C (i, j, X).

Count how many pairs of integers (i, j) meet the following conditions:

* 1 \ leq i \ leq N
* 1 \ leq j \ leq M
* C (i, j, A) \ neq C (i, j, B)



Input format


N M
A_1_1
...
A_N
B_1_1
...
B_N


When the jth character of A_i (1 \ leq i \ leq N) is `#`, C (i, j, A) is black, and when it is `.`, C (i, j, A) is white. Represents that. The same is true for B.

Constraint

* 1 \ leq N, M \ leq 2,000
* | A_i | = | B_i | = M
* A_i and B_i are strings consisting only of `#` and `.`



Output format

Print the answer on one line.

Input example 1


twenty three
.. #
..
.##
..


Output example 1


2

Input example 2


7 28
............................
... # ..... ### ... #### .... ### ..
.. #. # ... # ... # ... # ... # .... # ... #.
. # ... # ... # ...... #### ... # .....
. ##### .. # ... # .. # ...... # ... #.
. # ... # ... ### ... # ....... ### ..
............................
............................
.. ### .... ### ..... ## .... ### ..
. # ... # ... # ... # ... #. # ... # ... #.
...##...#...# ..... # .... ####.
.. # ..... # ... # ..... # ....... #.
. ##### ... ### .... #### ... ### ..
............................


Output example 2


40





Example

Input

2 3
..#
##.
.##
#..


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define whlie while
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define rep(i,N) for(int i = 0; i < (N); i++)
#define repr(i,N) for(int i = (N) - 1; i >= 0; i--)
#define rep1(i,N) for(int i = 1; i <= (N) ; i++)
#define repr1(i,N) for(int i = (N) ; i > 0 ; i--)
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define vrep(v,it) for(auto it = v.begin(); it != v.end(); it++)
#define vrepr(v,it) for(auto it = v.rbegin(); it != v.rend(); it++)
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) ll __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
using namespace std; void solve();
using ll = long long; using vl = vector<ll>;
using vi = vector<int>; using vvi = vector< vector<int> >;
constexpr int inf = 1001001001;
constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); rep(i,s) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}
#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,...) do { cerr << #v << " = "; array_out(v , __VA_ARGS__ );} while(0)
#else
  #define trc(...)
  #define trca(...)
  int main(){solve();}
#endif
 
using P = pair<int,int>; using vp = vector<P>;
constexpr int MOD = /**/ 1000000007; //*/ 998244353;
/////////////////

void solve(){
  ini(H,W);
  vector<string> A(H),B(H);
  in(A); in(B);
  int cnt = 0;
  rep(i,H) rep(j,W) cnt += A[i][j] != B[i][j];
  out(cnt);
}