import std; void main () { int N, M; readln.read(N, M); auto e = readln.split.to!(int[]); auto v = new int[](M); auto w = new int[](M); foreach (i; 0..M) { readln.read(v[i], w[i]); } // 複数のナップサックに適切な部分集合を割り当てていく -> 例のO(3^M)チャンス auto cost = new long[](1 << M); auto value = new long[](1 << M); foreach (S; 0..(1 << M)) { long co = 0, va = 0; foreach (i; 0..M) if (0 < (S & (1 << i))) { co += w[i]; va += v[i]; } cost[S] = co; value[S] = va; } auto dp = new long[][](N + 1, 1 << M); // dp[i][S] := 最初のi個のナップサックに割り当てる部分集合を確定させて、未使用アイテムがSの状態における最大価値 auto prev = new int[][](N + 1, 1 << M); foreach (p; prev) p[] = -1; foreach (d; dp) d[] = -long.max; dp[0][(1 << M) - 1] = 0; foreach (i; 0..N) { foreach (S; 0..(1 << M)) { if (dp[i][S] == -long.max) continue; // 非空部分集合の列挙 for (uint sub = S; 0 < sub; sub = (sub - 1) & S) { if (cost[sub] <= e[i]) { if (dp[i + 1][S ^ sub] < dp[i][S] + value[sub]) { dp[i + 1][S ^ sub] = dp[i][S] + value[sub]; prev[i + 1][S ^ sub] = sub; } } } // 取らない if (dp[i + 1][S] < dp[i][S]) { dp[i + 1][S] = dp[i][S]; prev[i + 1][S] = 0; } } } long ans = 0; foreach (d; dp[N]) ans = max(ans, d); writeln(ans); // 復元 auto took = new int[](0); int cur = N; int set = () { foreach (S; 0..(1 << M)) if (dp[N][S] == ans) { return S; } assert(0); }(); while (0 < cur) { took ~= prev[cur][set]; set |= prev[cur][set]; cur--; } took.reverse; foreach (t; took) { import core.bitop : popcnt; write(popcnt(t)); foreach (i; 0..M) if (0 < (t & (1 << i))) { write(" ", i + 1); } write("\n"); } } void read (T...) (string S, ref T args) { import std.conv : to; import std.array : split; auto buf = S.split; foreach (i, ref arg; args) { arg = buf[i].to!(typeof(arg)); } }