#include <bits/stdc++.h>
namespace MX {
constexpr int A = 26;
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
// consider string as a number in modulo 26
//
// call string a, b the 2 strings you need to check if close (after the op)
// clarification in first example (abc is string a, ced is string b)
//
// doing op (adding 11 to the pair) on string b is equivalent to subtracting
// string a, so we can keep string b constant
//
// we know how much to add on each position, consider an array that
// represents value needed to add to char in string a to get b start with 0,
// and construct bi - ai via the +1 on consec operation (and its reverse as
// mentioned above) for which array is it possible to build such an array so
// when added to ai you get bi
//
// lets call this array above c, ci = bi - ai for all i
//
// check selected problems: https://oj.uz/problem/view/COI23_bliskost for
// diagram
//
// so basically first operation is c_0, c_1 - c_0, c2 - c1 + c0 ... c_n-2 -
// c_n-3 + ... +/- c0 should be equal to c_(n - 1) ofc in modulo 26
//
// c_(n - 1) + c_(n - 3) + ... = c_(n - 2) + c_(n - 4) + ...
//
// so odd positions sum should be equal to even positions sum (ofc in mod
// 26) whcih we can keep track trivially
int N, Q;
std::cin >> N >> Q;
std::string A;
std::cin >> A;
std::string B;
std::cin >> B;
int odd = 0, eve = 0;
for (int i = 0; i < N; i++) {
(i % 2 ? odd : eve) += (B[i] - A[i] + MX::A) % MX::A;
odd += MX::A;
odd %= MX::A;
eve += MX::A;
eve %= MX::A;
}
std::cout << (odd == eve ? "da" : "ne") << "\n";
for (int q = 0; q < Q; q++) {
int p;
char c;
std::cin >> p >> c;
p--;
(p % 2 ? odd : eve) -= (B[p] - A[p]);
A[p] = c;
(p % 2 ? odd : eve) += (B[p] - A[p]);
odd += MX::A;
odd %= MX::A;
eve += MX::A;
eve %= MX::A;
std::cout << (odd == eve ? "da" : "ne") << "\n";
}
return 0;
}
