An optimal and efficient solution in Go that solves the problem by simulating the greedy interval covering strategy on the union of the given segments.

```go
package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

func main() {
	// Configure fast I/O
	scanner := bufio.NewScanner(os.Stdin)
	scanner.Split(bufio.ScanWords)
	writer := bufio.NewWriter(os.Stdout)
	defer writer.Flush()

	// Helper to read next integer
	nextInt := func() int {
		scanner.Scan()
		val, _ := strconv.Atoi(scanner.Text())
		return val
	}

	// Read n and m
	if !scanner.Scan() {
		return
	}
	n, _ := strconv.Atoi(scanner.Text())
	
	if !scanner.Scan() {
		return
	}
	m, _ := strconv.Atoi(scanner.Text())

	// Data structures
	type Interval struct {
		R  int
		ID int
	}

	// starts[i] stores the interval starting at i with the maximum reach (R)
	starts := make([]Interval, n+2)
	// diff array for computing union coverage
	diff := make([]int, n+2)

	for i := 1; i <= m; i++ {
		a := nextInt()
		l := nextInt()
		r := a + l - 1

		// Keep the interval with largest R for each start position a
		if r > starts[a].R {
			starts[a] = Interval{R: r, ID: i}
		}

		// Update difference array
		diff[a]++
		if r+1 <= n {
			diff[r+1]--
		}
	}

	// Determine which cells are covered by at least one instruction
	covered := make([]bool, n+2)
	currentDepth := 0
	for i := 1; i <= n; i++ {
		currentDepth += diff[i]
		if currentDepth > 0 {
			covered[i] = true
		}
	}

	kept := make([]bool, m+1)
	ptr := 1 // Pointer for scanning starts array

	// Iterate through memory to process connected components
	for i := 1; i <= n; {
		if !covered[i] {
			i++
			continue
		}

		// Identify the continuous component [S, E]
		S := i
		E := i
		for E < n && covered[E+1] {
			E++
		}

		// Solve interval covering for [S, E]
		curr := S - 1
		
		// Ensure scanning pointer is at least at start of component
		if ptr < S {
			ptr = S
		}

		// Track the best interval (furthest reach) found in the current scan window
		bestR := -1
		bestID := -1

		// Greedily extend coverage until E is reached
		for curr < E {
			limit := curr + 1 // We must cover the next byte (curr + 1)
			
			// Scan all intervals starting <= limit to find the one that reaches furthest
			for ptr <= n && ptr <= limit {
				if starts[ptr].R > bestR {
					bestR = starts[ptr].R
					bestID = starts[ptr].ID
				}
				ptr++
			}
			
			// Select the best interval found
			// Note: A solution is guaranteed to exist because [S, E] is covered in the union
			kept[bestID] = true
			curr = bestR
		}
		
		// Move to next possible component
		i = E + 1
	}

	// Collect removed instructions
	removed := make([]int, 0, m)
	for i := 1; i <= m; i++ {
		if !kept[i] {
			removed = append(removed, i)
		}
	}

	// Output results
	fmt.Fprintln(writer, len(removed))
	for i, id := range removed {
		if i > 0 {
			fmt.Fprint(writer, " ")
		}
		fmt.Fprint(writer, id)
	}
	fmt.Fprintln(writer)
}
```