#include <bits/stdc++.h>

// A の累積和を求める前計算
[[nodiscard]] static inline constexpr std::vector<uint_least64_t> prepare(const uint_fast32_t M, const std::vector<uint_least32_t>& A) noexcept
{
	std::vector<uint_least64_t> csum_A(M + 1, 0);
	for (uint_fast32_t i = 0; i != M; ++i)
		csum_A[i + 1] = csum_A[i] + A[i];

	return csum_A;
}

[[nodiscard]] static inline constexpr std::vector<std::pair<uint_least32_t, uint_least32_t>> solve([[maybe_unused]] const uint_fast32_t N, const uint_fast32_t M, const std::vector<uint_least32_t>& A) noexcept
{
	const std::vector<uint_least64_t>& csum_A = prepare(M, A);
	std::vector<std::pair<uint_least32_t, uint_least32_t>> ans;
	ans.reserve(M);
	for (uint_fast32_t i = 0, l = 1, r = M + 1; i != M; i = l, l = i + 1, r = M + 1)
	{
		while (l + 1 < r)  // 二分探索
		{
			const uint_fast32_t c = (l + r) / 2;
			if (csum_A[c] - csum_A[i] == static_cast<uint_fast64_t>(c - i) * (A[i] * 2 + (c - i - 1)) / 2)  // 右辺は等差数列の総和の公式の公差 1 の場合
				l = c;
			else
				r = c;
		}
		ans.emplace_back(A[i], l - i);
	}

	return ans;
}

static inline void output(const std::vector<std::pair<uint_least32_t, uint_least32_t>>& ans) noexcept
{
	std::cout << ans.size() << '\n';
	for (const auto& [S, L] : ans)
		std::cout << S << ' ' << L << '\n';
}

int main()
{
	std::cin.tie(nullptr);
	std::ios::sync_with_stdio(false);
	
	uint_fast32_t N, M;
	std::cin >> N >> M;
	std::vector<uint_least32_t> A(M);
	for (auto& a : A) std::cin >> a;

	output(solve(N, M, A));
	return 0;
}