#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<int> build_next(const string& p) {
    int n = p.size();                 // 模式串长度
    vector<int> nxt(n, 0);            // 存放 next 数组，初始化全 0

    for (int i = 1, j = 0; i < n; i++) {
        // --- step1: 如果当前字符不匹配，回退 j ---
        while (j > 0 && p[i] != p[j]) 
            j = nxt[j - 1];           // j 回退到上一个最长前后缀的长度，采用动态规划思想

        // --- step2: 如果匹配成功，j 前进 ---
        if (p[i] == p[j]) 
            j++;

        // --- step3: 记录当前前缀的 next 值 ---
        nxt[i] = j;                   // 表示 p[0..i] 的最长公共前后缀长度
    }

    return nxt;
}


vector<int> next_arr(const string& x) {
    int n = x.size();
    vector<int> tmp(n, 0);

    for (int i = 0; i < n; i++) {       
        int j = 0;
        for (int len = i; len > 0; len--) {
            if (x.substr(0, len) == x.substr(i - len + 1, len)) {
                j = len;
                break;
            }
        }
        tmp[i] = j;
    }
    return tmp;
}


int main(){
    string mod, input;
    cin >> input >> mod;
    vector<int> next = build_next(mod);
    int n = mod.size();
    int pi = 0, pm = 0;
    cout << next[n / 4] << " " << next[n / 2] << " " << next[3 * n / 4] << endl;
    while (pi < input.size()) {
        if (input[pi] == mod[pm]) {
            pi++;
            pm++;
            if (pm == mod.size()) {
                cout << pi - pm << " ";   // 匹配位置（起始下标）
                return 0;
                pm = next[pm - 1];       // 回退
            }
        } else {
            if (pm > 0)
                pm = next[pm - 1];       // 回退
            else
                pi++;
        }
    }
    cout << -1 << endl;
    return 0;
}

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