Problem: Example

Input

2 1
WE


Output

1 2

Solution (in CPP):

Solution:
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <functional>
#include <numeric>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <fstream>
#include <bitset>
#include <time.h>
#include <tuple>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<ll> V;
typedef complex<double> Point;

#define PI acos(-1.0)
#define EPS 1e-10
const ll INF = 1e12;
const ll MOD = 1e9 + 7;

#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define rep(i,N) for(ll i=0;i<(N);i++)
#define ALL(s) (s).begin(),(s).end()
#define EQ(a,b) (abs((a)-(b))<EPS)
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
#define fi first
#define se second
#define N_SIZE (1LL << 20)
#define NIL -1
#define MAX_N 100100

ll sq(ll num) { return num*num; }
ll mod_pow(ll x, ll n) {
	if (n == 0)return 1;
	if (n == 1)return x%MOD;
	ll res = sq(mod_pow(x, n / 2));
	res %= MOD;
	if (n % 2 == 1) {
		res *= x;
		res %= MOD;
	}
	return res;
}
ll mod_add(ll a, ll b) { return (a + b) % MOD; }
ll mod_sub(ll a, ll b) { return (a - b + MOD) % MOD; }
ll mod_mul(ll a, ll b) { return a*b % MOD; }

int n, m;
string s[110];
int cnt[110];

int main() {
	cin >> n >> m;
	ll e = 0;
	rep(i, m) {
		cin >> s[i];
		rep(j, n) {
			cnt[i] += (s[i][j] == 'W');
		}
		e += cnt[i];
	}
	ll ans = 0;
	rep(i, n) {
		ll sum = 0;
		rep(j, m) {
			if (s[j][i] == 'W')cnt[j]--;
			else cnt[j]++;
			sum += cnt[j];
		}
		if (sum < e) {
			e = sum;
			ans = i + 1;
		}
	}
	cout << ans << " " << ans + 1 << endl;
}